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Tristan Needham Visual Differential Geometry,pg-32

In the beginning, I understood the metric as the factors by which the length of displacement on surface and on the plane relate. But, in the book the following formula is given and suggests to me that separation vectors on the surface are related to separation vectors in the plane:

$$ d \hat{s} = \lambda(z, \gamma)dz$$

With, $dz = e^{i \gamma} ds$

My doubt is that the separation vector on the surface is existing in the tangent plane at the base point $\hat{z}$ and is three dimensional, so how could it possibly be related to the complex separation vector $dz$ existing in the flat plane?

tryst with freedom
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1 Answers1

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$\newcommand{\Cpx}{\mathbf{C}}$The formula $$ d\hat{s} = \lambda(z, \gamma)\, dz $$ must be interpreted not as equating a spatial displacement with a scalar multiple of the "flat" displacement $dz$, but as an equality between an infinitesimal displacement $d\hat{s} = \hat{q} - \hat{z}$ in the tangent plane to the hemisphere at $\hat{z}$ and the image of the "flat" displacement $dz = q - z$ under the differential of (the inverse of) central projection from the plane to the hemisphere.

More precisely, if $F:\Cpx \to H$ denotes the inverse projection from the plane to the open hemisphere, then $\hat{z} = F(z)$ and (speaking infinitesimally) $$ \hat{z} + d\hat{s} = \hat{q} = F(q) = F(z + dz) = F(z) + (DF(z))(dz); $$ consequently, $d\hat{s} = (DF(z))(dz)$.

(The identification of $(DF(z))(dz)$ with a scalar multiple of $dz$ presumably represents the differential with respect to some basis.)

Andrew D. Hwang
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  • Could you explain in a less formal way? The answer's language is a bit difficult to understand for one who doesn't understand the technical terms completely – tryst with freedom Nov 28 '21 at 14:45
  • Qualitatively, if $\hat{q} = \hat{z}+d\hat{s}$ moves on the sphere, then $q=z+dz$ moves in the plane, and vice versa. Needham's formula specifies how to relate a change in $q$ with the corresponding change in $\hat{q}$. Arguably the notation is not ideal, however, for the reasons giving rise to your question: There is an underlying identification between two non-parallel planes that is concealed by the notation. – Andrew D. Hwang Nov 28 '21 at 15:58
  • Thanks, the last part made it more clear – tryst with freedom Nov 28 '21 at 16:53
  • To double check, we are taking the tangent plane at the point and seeing what displacement is induced there by motion in the map right? – tryst with freedom Nov 28 '21 at 16:58
  • Yes, your previous comment is right, modulo misunderstanding on my part from not having read the book. – Andrew D. Hwang Nov 28 '21 at 17:09